# Finding a primary decomposition

Let $k$ be a field, and $R=k[x,y]$.

I'm supposed to find two different minimal primary decompositions of the ideal $(x^2y, y^2x)$.

It's easy to see that one minimal primary decomposition is $(x)\cap(y)\cap(x^2, y^2)$.

My question: What's the second minimal primary decomposition of the above ideal?

EDIT: I've now added the minimality requirement to the question.

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Is it necessary that they be minimal? – Qiaochu Yuan May 23 '11 at 19:06
For some reason I can't edit my original question, so here is an important change (which answers Qiaochu's question): Yes, the decompositions have to be minimal. – user11281 May 23 '11 at 19:32
Your accounts have been merged. Since your new account is registered you should no longer have problems with logging in. – Qiaochu Yuan May 23 '11 at 19:53

Since you suggest this is homework, I'll put this in the form of hints:

(1) Remember that the set of associated primes for a primary decomposition is unique. So you are looking for $\mathfrak{p} \cap \mathfrak{q} \cap \mathfrak{r}$ where $\sqrt{\mathfrak{p}} = (x)$, $\sqrt{\mathfrak{p}} = (y)$ and $\sqrt{\mathfrak{r}} = (x,y)$.

(2) Show that $\mathfrak{p}$ and $\mathfrak{q}$ must be $(x)$ and $(y)$. So the place where you have room to play is in choosing $\mathfrak{r}$.

At this point, I have trouble giving good general hints. The next two, which are far more helpful, will be put in ROT13.

(3) Lbh jvyy abg or noyr gb fbyir guvf ceboyrz vs lbh fgvpx gb vqrnyf trarengrq ol zbabzvnyf.

(4) Gur rknzcyr V sbhaq jnf nyfb trarengrq va qrterr gjb.

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I'm a bit late with this but even so:

\begin{align} (x^2 y, y^2 x) &= (x^2 y - y^2 x, y^2 x) \\ &= (xy(x-y), y^2 x) \\ &= (y) \cap (x(x-y), y^2x) \\ &= (y) \cap (x) \cap (x-y, y^2) \end{align}

It remains to be shown that $(x-y, y^2)$ is primary:

Let $r(I)$ denote the radical of $I$. Then $r((x-y, y^2)) = (x-y, y) = (x,y)$. To see that $(x,y)$ is maximal note that $k[x,y]/(x,y) \cong k$ is a field. We know that if $r(I)$ is maximal then $I$ is primary so we're done.

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